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    #21 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    I didn't check them all, but indeed it can't be right that $2$ is in this list twice, nor that it maps to two different elements.
    Why is it in the list twice?
    We start from the right-most cycle, right? Then we have the element $2$ and see that this goes to $4$ and $4$ goes to $9$, so $2$ goes to $9$.
    At the second cycle we have again the element $2$. When we reach this cycle do we check again the element $2$ ?

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    #22
    Quote Originally Posted by mathmari View Post
    We start from the right-most cycle, right? Then we have the element $2$ and see that this goes to $4$ and $4$ goes to $9$, so $2$ goes to $9$.
    At the second cycle we have again the element $2$. When we reach this cycle do we check again the element $2$ ?
    No.
    When we consider the image of an element such as $2$, we must evaluate all cycles from right to left. We are not supposed to start in the middle.

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    #23 Thread Author
    Ok !!

    Now I think I got the right results :

    $\pi_4=\pi_1\circ\pi_2=(14\ 13) \ (6\ 3\ 2\ 9\ 1)\ (12\ 8\ 7\ 4\ 5)$

    $\pi_5=\pi_2\circ\pi_3=(1\ 4) \ (2\ 8\ 11\ 12\ 6\ 7)\ (13\ 14)$

    $\pi_6=\pi_2\circ\pi_1=(3 \ 4 \ 9 \ 12\ 1) \ (2\ 5\ 8\ 6\ 7)\ (13 \ 14)$

    For the signum we have to write each permutation as a product of cycles of two elements, right? How can we do that?

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    #24
    Quote Originally Posted by mathmari View Post
    For the signum we have to write each permutation as a product of cycles of two elements, right? How can we do that?
    We can write $(a\,b\,c\,d)$ for instance as either $(a\,d)(a\,c)(a\,b)$ or $(a\,b)(b\,c)(c\,d)$.

    Note that more generally a cycle with an even number of elements can be written as an odd number of 2-cycles (signum -1).
    And a cycle with an odd number of elements can be written as an even number of 2-cycles (signum +1).

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    #25 Thread Author
    I am looking aagain at the previous question, at the calculation of $\pi_4$. Do we get the same results if we use the initially given $\pi_1$ and $\pi_2$ and the same if we use the cycle decomposition of them?

    I get the following from the cycle decomposition:

    $\pi_1=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 )$

    $\pi_2=(7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$

    $\pi_4=\pi_1\circ\pi_2=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 ) \ (7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$

    $14\rightarrow 13$

    $13\rightarrow 14$

    $6\rightarrow 12\rightarrow 8$

    $12\rightarrow 1\rightarrow 9$

    $1\rightarrow 6\rightarrow 12$

    $7\rightarrow 8\rightarrow 6$

    $8\rightarrow 5$

    $5\rightarrow 4\rightarrow 2$

    $4\rightarrow 2\rightarrow 3$

    $2\rightarrow 7$

    $9\rightarrow 4$

    $3\rightarrow 1$

    So $\pi_4=(14 \ 13) \ (6 \ 8 \ 5 \ 2 \ 7) \ (12 \ 9 \ 4 \ 3 \ 1) $.



    And from the initially given forms:

    $\pi_1=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12)$

    $\pi_2=(2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 14)$

    $\pi_4=\pi_1\circ\pi_2=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12) \ (2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 14)$

    $13 \rightarrow 14$

    $14\rightarrow 13$

    $1\rightarrow 12 \rightarrow 6$

    $12\rightarrow 6\rightarrow 8$

    $6\rightarrow 1 \rightarrow 3$

    $2\rightarrow 4\rightarrow 9$

    $4\rightarrow 5$

    $5\rightarrow 8\rightarrow 12$

    $8\rightarrow 7$

    $7\rightarrow 2\rightarrow 4$

    $3\rightarrow 1\rightarrow 2$

    $9\rightarrow 1$

    So $\pi_4=(13 \ 14) \ (1 \ 6 \ 3 \ 2 \ 9 ) \ (12 \ 8 \ 7 \ 4 \ 5)$.


    Have I done something wrong or is it correct that we get different results?

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    #26
    The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it?

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    #27 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    The original $\pi_2$ was already disjoint and contained (1 12 6). But it's not like that in both your cases, is it?
    What do you mean? I got stuck right now.

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    #28
    Quote Originally Posted by mathmari View Post
    I get the following from the cycle decomposition:

    $\pi_1=(8 \ 6 \ 12)\ (1 \ 9 \ 4 \ 2 \ 3 )$

    $\pi_2=(7 \ 8 \ 5 \ 4 \ 2)\ (6 \ 12 \ 1 ) \ (14 \ 13)$
    You wrote (6 12 1) here.

    Quote Originally Posted by mathmari View Post
    And from the initially given forms:

    $\pi_1=(1 \ 2\ 4 \ 9)\ (1\ 3)\ (6 \ 8 \ 12)$

    $\pi_2=(2 \ 4\ 5 \ 8 \ 7)\ ( 1 \ 12 \ 6) \ (13 \ 1
    4)$
    And you wrote (1 12 6) here.
    Shouldn't they be the same?

    Did you perhaps accidentally write $\pi_2^{-1}$ instead of the decomposition?

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    #29 Thread Author
    Quote Originally Posted by Klaas van Aarsen View Post
    You wrote (6 12 1) here.



    And you wrote (1 12 6) here.
    Shouldn't they be the same?

    Did you perhaps accidentally write $\pi_2^{-1}$ instead of the decomposition?
    Oh yes, you are right! I used theinverse accidentaly.

    So we get the same result using the initial permutation and the cycle decomposition of it.

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